Oracle PL/SQL Tutorial/Regular Expressions Functions/REGEXP INSTR — различия между версиями

Материал из SQL эксперт
Перейти к: навигация, поиск
м (1 версия)
 
(нет различий)

Версия 13:45, 26 мая 2010

Find the "s" and ignore case.

SQL>
SQL> SELECT REGEXP_INSTR("This is a test for print SS","s",1,1,0,"i") position
  2  FROM dual;
  POSITION
----------
         4
SQL>


Occurrence refers to the first, second, third, etc., occurrence of the pattern in S. The default is 1 (first).

Searching for the second "a" starting at position 1.



SQL> SELECT REGEXP_INSTR("This is a test for print a","a",1,2) position
  2  FROM dual;
  POSITION
----------
        26
SQL>


Parameters is a field that may be used to define how one wants the search to proceed:

  1. i - to ignore case
  2. c - to match case
  3. n - to make the metacharacter dot symbol match new lines as well as other characters (more on this later in the chapter)
  4. m - to make the metacharacters ^ and $ match beginning and end of a line in a multiline string (more, later)

The default is "i".

The following example finds the "s" and match case.



SQL>
SQL> SELECT REGEXP_INSTR("This is a test for printing SS","s",1,1,0,"c") position
  2  FROM dual;
  POSITION
----------
         4
SQL>


Regexp_Instr

REGEXP_INSTR function returns a number for the position of matched pattern.

Unlike INSTR, REGEXP_INSTR cannot work from the end of the string backward.

The arguments for REGEXP_INSTR are:



REGEXP_INSTR(String to search, Pattern, [Position,[Occurrence, [Return-option, [Parameters]]]])


== regexp_instr(comments, "[^ ]+", 1,/td>




SQL>
SQL> create table history
  2  ( empno      NUMBER(4)
  3  , beginyear  NUMBER(4)
  4  , begindate  DATE
  5  , enddate    DATE
  6  , deptno     NUMBER(2)
  7  , msal       NUMBER(6,2)
  8  , comments   VARCHAR2(60)
  9  ) ;
Table created.
SQL>
SQL>
SQL> insert into history values (9,2000,date "2000-01-01",date "2002-01-02",40, 950,"history for 9");
1 row created.
SQL> insert into history values (8,2000,date "2000-01-02", NULL       ,20, 800,"");
1 row created.
SQL> insert into history values (7,1988,date "2000-01-06",date "2002-01-07",30,1000,"");
1 row created.
SQL> insert into history values (6,1989,date "2000-01-07",date "2002-01-12",30,1300,"");
1 row created.
SQL> insert into history values (5,1993,date "2000-01-12",date "2002-01-10",30,1500,"history for 5");
1 row created.
SQL> insert into history values (4,1995,date "2000-01-10",date "2002-01-11",30,1700,"");
1 row created.
SQL> insert into history values (3,1999,date "2000-01-11", NULL       ,30,1600,"");
1 row created.
SQL> insert into history values (2,1986,date "2000-01-10",date "2002-01-08",20,1000,"history for 2");
1 row created.
SQL> insert into history values (1,1987,date "2000-01-08",date "2002-01-01",30,1000,"history for 1");
1 row created.
SQL> insert into history values (7,1989,date "2000-01-01",date "2002-05-12",30,1150,"history for 7");
1 row created.
SQL>
SQL> select comments
  2  from   history
  3  where  regexp_instr(comments, "[^ ]+", 1, 9) > 0;
no rows selected
SQL>
SQL>
SQL> drop table history;
Table dropped.
SQL>
SQL>


Regexp_Instr returns the location (beginning) of a pattern in a given string

REGEXP_INSTR extends the regular INSTR string function by allowing searches of regular expressions.

The simplest form of this function is:



REGEXP_INSTR(source_string, pattern_to_find)


REGEXP_INSTR(x, pattern [, start [, occurrence [, return_option [, match_option]]]]) searches for pattern in x.

REGEXP_INSTR() returns the position at which pattern occurs.

The position starts at number 1.

The following example returns the position that matches the regular expression lalpha: {4} using REGEXP_INSTR():



SQL> SELECT REGEXP_INSTR("abcedfghijklumnoprstuvwxyzabcedfghijklumnoprstuvwxyzabcedfghijklumnoprstuvwxyz", "l[[:alpha:]]{4}") AS result
  2  FROM dual;
    RESULT
----------
        12
SQL>


Return-option returns the position of the start or end of the matched string.

The default is 0, which returns the starting position of the pattern in the target.

A value of 1 returns the starting position of the next character following the pattern match.



SQL>
SQL> SELECT REGEXP_INSTR("This is a test","a",1,2,0) position
  2  FROM dual;
  POSITION
----------
         0
SQL>


Returns the position of the second occurrence that matches the letter o starting at position 10 using REGEXP_INSTR()

SQL>
SQL> SELECT REGEXP_INSTR("abcedfghijklumnoprstuvwxyzabcedfghijklumnoprstuvwxyzabcedfghijklumnoprstuvwxyz", "o", 10, 2) AS result
  2  FROM dual;
    RESULT
----------
        42
SQL>


Returns the position of the second occurrence that matches the regular expression salpha:{3} starting at position 1:

SQL>
SQL> SELECT REGEXP_INSTR("But, soft! What light through yonder window softly breaks?", "s[[:alpha:]]{3}", 1, 2) AS result
  2  FROM dual;
    RESULT
----------
        45
SQL>


SELECT REGEXP_INSTR("Two is bigger than One","One") where_it_is

2  FROM dual;
WHERE_IT_IS
-----------
         20
SQL>


Specify any series of letters and find matches, just like INSTR

SQL>
SQL> -- create demo table
SQL> create table myTable(
  2    id           NUMBER(2),
  3    value        VARCHAR(50)
  4  );
Table created.
SQL>
SQL> insert into myTable(id, value)values(1,"1234 4th St. Vancouver");
1 row created.
SQL> insert into myTable(id, value)values(2,"4 Maple Ct. New York");
1 row created.
SQL> insert into myTable(id, value)values(3,"4321 Green Blvd. London");
1 row created.
SQL> insert into myTable(id, value)values(4,"33 Third St. Toronto");
1 row created.
SQL> insert into myTable(id, value)values(5,"One First Drive. Queen");
1 row created.
SQL> insert into myTable(id, value)values(6,"1664 1/2 Springhill Ave");
1 row created.
SQL> insert into myTable(id, value)values(7,"665 Fall Ave. Linken");
1 row created.
SQL>
SQL> select * from mytable;
        ID VALUE
---------- --------------------------------------------------
         1 1234 4th St. Vancouver
         2 4 Maple Ct. New York
         3 4321 Green Blvd. London
         4 33 Third St. Toronto
         5 One First Drive. Queen
         6 1664 1/2 Springhill Ave
         7 665 Fall Ave. Linken
7 rows selected.
SQL>
SQL>
SQL> SELECT value, REGEXP_INSTR(value,"ing") where_it_is
  2  FROM myTable
  3  WHERE REGEXP_INSTR(value,"ing") > 0;
VALUE                                              WHERE_IT_IS
-------------------------------------------------- -----------
1664 1/2 Springhill Ave                                     13
SQL>
SQL>
SQL> drop table myTable;
Table dropped.
SQL>
SQL>


The Return-option is set to 1 to indicate the end of the found pattern

SQL>
SQL> SELECT REGEXP_INSTR("This is a test","a",1,2,1) position
  2  FROM dual;
  POSITION
----------
         0
SQL>


The simplest regular expression matches letters, letter for letter

SQL>
SQL> -- create demo table
SQL> create table myTable(
  2    id           NUMBER(2),
  3    value        VARCHAR(50)
  4  );
Table created.
SQL>
SQL> insert into myTable(id, value)values(1,"1234 4th St. Vancouver");
1 row created.
SQL> insert into myTable(id, value)values(2,"4 Maple Ct. New York");
1 row created.
SQL> insert into myTable(id, value)values(3,"4321 Green Blvd. London");
1 row created.
SQL> insert into myTable(id, value)values(4,"33 Third St. Toronto");
1 row created.
SQL> insert into myTable(id, value)values(5,"One First Drive. Queen");
1 row created.
SQL> insert into myTable(id, value)values(6,"1664 1/2 AAA Ave");
1 row created.
SQL> insert into myTable(id, value)values(7,"665 Fall Ave. Linken");
1 row created.
SQL>
SQL> select * from mytable;
        ID VALUE
---------- --------------------------------------------------
         1 1234 4th St. Vancouver
         2 4 Maple Ct. New York
         3 4321 Green Blvd. London
         4 33 Third St. Toronto
         5 One First Drive. Queen
         6 1664 1/2 AAA Ave
         7 665 Fall Ave. Linken
7 rows selected.
SQL>
SQL> SELECT value, REGEXP_INSTR(value,"Ave") where_it_is
  2  FROM myTable
  3  WHERE REGEXP_INSTR(value,"Ave") > 0;
VALUE                                              WHERE_IT_IS
-------------------------------------------------- -----------
1664 1/2 AAA Ave                                            14
665 Fall Ave. Linken                                        10
SQL>
SQL> drop table myTable;
Table dropped.
SQL>
SQL>